Question 14 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at Dec. 4, 2021 by Teachoo
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The tangent to the curve γy=eγ^2x at the point (0, 1) meets x-axis at:
(A)(0, 1)Β Β Β Β Β (B) β 1/2, 0
(C) (2, 0)Β Β Β Β Β (D) (0, 2)
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Question 14
The tangent to the curve γπ¦=πγ^2π₯ at the point (0, 1) meets x-axis at:
(0, 1) (B) β 1/2, 0
(C) (2, 0) (D) (0, 2)
Given
γπ¦=πγ^2π₯
Finding slope of tangent at (0, 1)
γπ¦=πγ^2π₯
Differentiating w.r.t. π₯
π π/π π = 2π^ππ
At point (0, 1)
Putting π₯=0, π¦=0 in ππ¦/ππ₯
ππ¦/ππ₯ = γ2πγ^0
ππ¦/ππ₯ = 2 Γ 1
π π/π π=π
Finding equation of tangent
Equation of line at (π₯1 , π¦1) & having Slope m is
π¦βπ¦1=π(π₯βπ₯1)
β΄ Equation of tangent at point (0, 1) & having slope 2 is
(πβπ) = 2 (πβπ)
(π¦β1) = 2π₯
π=ππ+π
Now, we need to find the point where tangent meets the x-axis
Since point is on the x-axis, itβs y-coordinate will be 0
β΄ Point = (π,π)
Putting π¦ = 0 in (1)
0 = 2π + 1
β1 = 2π₯
(β1)/2=π₯
π = (βπ)/π
Thus, required point is ((βπ)/π, π)
So, the Correct answer is (B)
Made by
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.
